A student suffering from myopia is not able to see distinctly the objects placed beyond 5 m. List two possible reasons due to which this defect of vision may have arisen. With the help of ray diagrams, explain 5 marks (i) why the student is unable to see distinctly the objects placed beyond 5 m from his eyes. (ii) The type of the corrective lens used to restore proper vision and how this defect is corrected by the use of this lens. (b) If, in this case, the numerical value of the focal length of the corrective lens is 5 m, find the power of the lens as per the new Cartesian sign convention.
Answers
(a) elongation of eyeball
(b) excessive curvature of the lens.
(i) student unable to see distinctly the objects placed beyond 5m from his eyes because student suffering with myopia . A myopic eye can't see distinct object clearly but able to see nearby object clearly.
(ii) convex lens of suitable focal length is used to restore proper vision. After using convex , image formed at ratina. While before it , image was formed near the ratina . As shown in figure it is clear how the lens correct this type of defect.
b. Focal length of convex lens is positive
So, f = +5m
Now, Power = 1/f = +1/5 = +0.2D
Hence, power is +0.2 diopter
Answer:
) Myopia is also known as near-sightedness. A person with myopia can see nearby objects clearly but cannot see distant objects distinctly. A person with this defect has the far point nearer than infinity. Such a person may see clearly upto a distance of a few metres. In a myopic eye, the image of a distant object is formed in front of the retina and not at the retina itself. This defect may arise due to: (11) excessive curvature of the eye lens, or (22) elongation of the eyeball.
(i) The student is able to see the objects clearly upto 5m5m, i.e., his far point is 5m5m. For a myopic eye with the near point as 5m5m, if any object is kept beyond this distance, the image is not formed on the retina. Instead, the image is formed in front of the retina as shown in the figure. Hence, the student is unable to see distinctly the object places beyond 5m5m from his eyes.
(ii) This defect can be corrected by using a concave lens of suitable power. A concave lens of suitable power will bring the image back on to the retina and thus the defect is corrected.
(b) Given, focal length (f) = -5m(f)=−5m
Power (P) = \cfrac { 1 }{ focal \ length (f) } = \cfrac { 1 }{ -5m } = -0.2 DPower(P)=
focal length (f)
1
=
−5m
1
=−0.2D
please mark brainliest